materials – powerpoint presentation 2015

54
Materials

Upload: tranthuy

Post on 27-Dec-2016

219 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Materials – PowerPoint Presentation 2015

Materials

Page 2: Materials – PowerPoint Presentation 2015

What is Materials Science and Engineering?

Structure

Performance

Processing

Properties

Materials EngineeringMaterials Science

Engineering Design

Page 3: Materials – PowerPoint Presentation 2015

The goal of materials science is to empower scientists and engineers to make informed decisions about the design, selection and use of materials for specific applications.

http://www.healthbase.com/resources/images/ortho/hip_replacement_implants/healthbase_zimmer_total_hip_replacement_implant_components.jpg

Photograph: Nils Jorgensen/REX

Under Armour piezoelectric adhttps://www.youtube.com/watch?v=MK7gYP9HWpY

Page 4: Materials – PowerPoint Presentation 2015

Four Fundamental Tenets Guide Materials Science

1. The principles governing the behavior of materials are grounded in science and are understandable

2. The properties of a given material are determined by its structure. Processing can alter the structure in specific and predictable ways

3. Properties of all materials change over time with use and exposure to environmental conditions

4. When selecting a material for a specific application, sufficient and appropriate testing must be performed to ensure that the material will remain suitable for its intended application throughout the intended life of the product

Page 5: Materials – PowerPoint Presentation 2015

A materials scientist or engineer must be able to:

1. Understand the properties associated with various classes of materials

2. Know why these properties exist and how they can be altered to make a material more suitable for a given application

3. Be able to measure important properties of materials and how those properties will impact performance

4. Evaluate the economic considerations that ultimately govern most material issues

5. Consider the long-term effects of using a material on the environment

Page 6: Materials – PowerPoint Presentation 2015

Fundamental types of materials important to engineering:

1. Crystals - Engineering metals and alloys Systemic, regular pattern, minimize volume

2. Engineering Ceramics (including glass)High viscosity at liquid-solid point prevents crystallization. These materials are usually amorphous

3. Polymers Long chains of simple, molecular structures. Plastics and living things

4. ElastomersLong chain polymers which fold or coil. Natural and artificial rubber. Enormous extensions associated with folding and unfolding of chains.

Page 7: Materials – PowerPoint Presentation 2015

Metals Ceramics

Polymers

http://www.buzzfeed.com/scott/nerd-venn-diagram

Semiconductors

Steel reinforced concrete

Concrete

Page 8: Materials – PowerPoint Presentation 2015

The Fundamental Material - Atom

1. What are the atomic building blocks?a) Nucleus – Protons (+) and Neutrons (0)b) Electrons (-)c) Atoms have a neutral charge (#protons =#electrons)

2. How are electrons distributed through an atom?a) Electrons in organized shells in an electron cloud b) # electrons/shell = 2N2 (N = the shell number)

3. What are valence electrons? Why are they important?a) Valence electrons are in the outermost shellb) Reactivity of atom depends upon valence electrons

4. Why are noble gases inert?a) The noble gases have full shells of electrons Atom = stadium

Nucleus = housefly in the center of the field

http://upload.wikimedia.org/wikipedia/commons/thumb/d/dc/M%26T_Bank_Stadium_DoD.jpg/300px-M%26T_Bank_Stadium_DoD.jpg

Page 9: Materials – PowerPoint Presentation 2015

Atoms – Can we see them?Electron Microscopy and Scanning Probe Microscopy

http://s.hswstatic.com/gif/atom-ibm.jpg

Xe on Ni

http://upload.wikimedia.org/wikipedia/commons/thumb/e/ec/Atomic_resolution_Au100.JPG/200px-Atomic_resolution_Au100.JPG

Au

Page 10: Materials – PowerPoint Presentation 2015

The Fundamental Material - Atom

1. All atoms of a given element are ______________

2. Atoms of different elements have different ________________

3. A compound is a specific combination of atoms of more than one element

4. In a chemical reaction, atoms are neither created nor destroyed – only change partners to produce new substances

HCl + NH3 NH4Cl

http://upload.wikimedia.org/wikipedia/commons/a/a0/Hydrochloric_acid_ammonia.jpg

Page 11: Materials – PowerPoint Presentation 2015

http://en.wikipedia.org/wiki/Periodic_table#mediaviewer/File:Periodic_table_(polyatomic).svg

Page 12: Materials – PowerPoint Presentation 2015

What holds the atoms in metals/crystals, ceramics, polymers and elastomers together?

BONDS!

Page 13: Materials – PowerPoint Presentation 2015

• Two or more atoms share electrons• Strong and rigid• Found in organics and sometimes ceramics• Strongly directional• Methane CH4

Carbon has ___ valence electronsHydrogen has ___ valence electrons

• Elemental solids – diamond• Can be stronger – diamond• Can be weaker - Bi

Covalent Bonds

Shared electron from hydrogen

Shared electron from carbon

http://myhome.sunyocc.edu/~weiskirl/methane2.GIF

Page 14: Materials – PowerPoint Presentation 2015

NO YES YES

Example: Na (+) (small) and Cl (-)(large)

Packing: as close as possible.

Ionic Bonds

• Bonding between a metal and a non-metal• Metal gives up valence electron(s) to non-metal• Results in all atoms having stable electron configuration• Na+Cl-

• Metal becomes +ly charged (cation); non-metal becomes –ly charged (anion)

• Coulombic attraction• Close-packed

http://upload.wikimedia.org/wikipedia/commons/a/ae/Sodium-chloride-3D-ionic-2.png

Page 15: Materials – PowerPoint Presentation 2015

Metallic Bonds

• Hold metals and alloys together• Allows for dense packing of atoms (why metals are heavy)• Valence electrons are not bound to a particular atom and

are free to drift through the entire material = “sea of electrons”

• Nonvalence electrons + atomic nuclei = ion core (with a net + charge)

• Good electrical conductivity• Good heat conductivity

Page 16: Materials – PowerPoint Presentation 2015

Bonds holding molecules together

Hydrogen bonds• Intermolecular attraction in which a H atom bonded to a small,

electronegative atom (N, O or F) is attracted to a lone pair of electrons on another N, O or F atom.

• Weak interactions• Due to the charge distribution on molecule• Often seen in organic compounds• 5 to 30 kJ/mole (as compared to about 100kJ/mole for chemical bond)

+ +

-

Example: H2O

Intermolecular Forces

-O-H…O=

Page 17: Materials – PowerPoint Presentation 2015

Bonds holding molecules together

Van der Waal forces:• Short-time interactions

• Arise from surface differences across molecules

• Weaker forces (~10 kJ/mole)

• Gecko feet: Microscopic branched elastic hairs on toes which take advantage of these atomic scale attractive forces to grip and support heavy loads

Intermolecular Forces

http://upload.wikimedia.org/wikipedia/en/0/03/Micro_and_nano_view_of_gecko's_toe.jpg

Page 18: Materials – PowerPoint Presentation 2015

Consequences of Structure

• Structure is related to the arrangement of the components of a material

• This could be on any length scale – atomic, nano-, micro-, macro-

• All length scales matter

• Types of carbon (literally just carbon!)a) Diamondb) Graphitec) Lonsdeleited) Buckminster Fullerene C60e) C540f) C70g) Amorphous carbonh) Carbon Nanotube

http://i17.photobucket.com/albums/b66/BuddyChrist80/Science/Chemistry/Eight_Allotropes_of_Carbon.png

Page 19: Materials – PowerPoint Presentation 2015

Spaghetti!

Take a 10 minute break

During the break take time to think about spaghettiHow does it break?How many ways can it break?How can it be made stronger?

Page 20: Materials – PowerPoint Presentation 2015

Material Properties

Young’s ModulusTensile StrengthYield StrengthCompressive StrengthShear StrengthDuctilityPoisson’s RatioSpecific WeightSpecific ModulusHardness

Stress-Strain CurveWe will determine for Spaghetti

Page 21: Materials – PowerPoint Presentation 2015

Mechanical Properties

How easy does spaghetti break in tension (by pulling)?Is thicker spaghetti easier or harder to break in tension?

Theory says that force needed to break in tension increases with cross-sectional area.

How easily does spaghetti buckle in compression?Depends on force, material strength, length and thickness of spaghetti

A longer piece buckles easier than a shorter pieceA thinner piece buckles easier than a thicker piece

How easily does spaghetti bend if you push on it perpendicularly?Is it in tension or compression?

Deflection depends on force, material strength, span length and cross-sectional area.A larger force yields a larger deflectionFor a given force, longer pieces bend easierFor a given force, thin pieces bend easier

Page 22: Materials – PowerPoint Presentation 2015

Terms associated with material properties

Hardness --resistance to scratching and denting.

Malleability --ability to deform under rolling or hammering without fracture.

Toughness --ability to absorb energy, e.g., a blow from a hammer. Area under stress-strain curve is a measure of toughness.

Ductility --ability to deform under tensile load without rupture; high percentage elongation and percent reduction of area indicate ductility

Brittleness --material failure with little deformation; low percent elongation and percent area reduction.

Elasticity --ability to return to original shape and size when unloaded

Plasticity --ability to deform non-elastically without rupture

Stiffness --ability to resist deformation; proportional to Young’s Modulus E (psi) E = stress/strain (slope of linear portion of stress/strain curve).

Page 23: Materials – PowerPoint Presentation 2015

Stress and Strain

Stress is related to the force or load applied to a material

Stress = σ = Force/original areaFrom Figure: σ = F/A0

Force - units? Newton = kg*m/s2

σ = F/A0 units = N/m2 = Pascal

MPa = 106 PaGPa = 109 Pa

F

F

0

0

Page 24: Materials – PowerPoint Presentation 2015

Stress and StrainStrainStrain = ε = change in length divided by original length

From Figure: ε = Δ / 0

Units? Strain is unitlessMay be reported as:

m/m, in/in or %

Elastic Strain vs. Plastic Strain(Rubber Band vs. Silly Putty)

Elastic--loading and unloading returns material to original length-can be done repeatedly, e.g., a watch spring.

Plastic--larger deformations are not reversible when "elastic limit" is exceeded. Some materials are almost purelyplastic, e.g., putty.

F

F

0

Page 25: Materials – PowerPoint Presentation 2015

Hooke's LawRobert Hooke, 1679 “As the extension, so the force”

i.e., stress is proportional to strain

Hooke's Law: an approximation of the relationship between the deformation of molecules and interatomic forces.

interatomic

distance

force

(tension)

neutral position

http://upload.wikimedia.org/wikipedia/commons/1/10/13_Portrait_of_Robert_Hooke.JPG

Page 26: Materials – PowerPoint Presentation 2015

Hooke’s Law

Hooke’s Law of Spring Extension

F=k*x

Hooke’s Law applied to Linear Elastic Solids

σ = E*ε

Thomas Young (1773 – 1829)

http://upload.wikimedia.org/wikipedia/commons/9/9f/Thomas_Young_(scientist).jpg

Young’s Modulus• The ratio of stress to strain within the linear

(elastic) region of the stress-strain curve• A measure of the “stiffness” of a material• Also known as the Modulus of Elasticity• Units are the same as the units of stress (F/A)

E = σ/ε

Page 27: Materials – PowerPoint Presentation 2015

Elastic solids – Young’s Modulus

Think of E as the stress required to deform a solid by 100%. (Most solids will fail at an extension of about 1%, so this is usually hypothetical).

Range of E in materials is enormous:

E(metal) 45-400 GPaE(Ceramics) 60-500 GPaE(Polymers) 0.01-4 GPaE(Spaghetti) 4.8 Gpa

Higher E implies? Greater Stiffness

Load

Unload

Slope = Young’s Modulus

Strain

Stre

ss (

Pa)

Page 28: Materials – PowerPoint Presentation 2015

Material testing – Tensile Strength

Usually tested by controlling extension (strain) and measuring resulting load (stress*area), i.e., independent variable is strain, dependent variable is stress

Can also be determined by subjecting material to a predetermined load and measuring elongation, i.e., independent variable is stress, dependent variable is strain

Page 29: Materials – PowerPoint Presentation 2015

Solid Behavior - Tension

After tensile testing:

a) Brittle

b) Ductile

c) Completely Ductile

Examples:

a) Cast Iron

b) Aluminum

c) Putty*

Page 30: Materials – PowerPoint Presentation 2015

Stress-Strain Curves

Str

ess,

s (

MP

a)

Strain, e (m/m)

Ds

De

E

11. Proportionality Limit

2. Elastic Limit

3. Yield Strength

4. Ultimate Tensl Strgth

5. Fracture Strength

E = Young’s Modulus,

Modulus of Elasticity

Elastic Region

Plastic Region

2

3

Tensile Test

45

E

0.2%

Page 31: Materials – PowerPoint Presentation 2015

Range of Tensile Strengths (TS) - How hard

a pull is required to break material bonds?

steel piano wire = 450,000 p.s.i.

aluminum = 10,000 p.s.i.

concrete = 600 p.s.i.

Page 32: Materials – PowerPoint Presentation 2015

Which curve is typical of:A ductile materialA brittle material

Page 33: Materials – PowerPoint Presentation 2015

Stress – Strain Curves

Page 34: Materials – PowerPoint Presentation 2015

Young’s Modulus

Page 35: Materials – PowerPoint Presentation 2015

Material Strength - Compression

Materials fail in compression in many ways depending on their geometry and support

a) buckling--hollow cylinders, e.g., tin can

b) bending--long rod or panel

c) shattering--heavily loaded glass

No relation between compressive and tensile strength in part because distinction between a material and a structure is often not clear. e.g., what is a brick? or concrete?

Page 36: Materials – PowerPoint Presentation 2015

Compressive strength of material

Under compression a beam will fail either by crushing or buckling, depending on the material and L/d; e.g., wood will crush if L/d < 10 and will buckle if L/d > 10 (approximately).

Crushing: atomic bonds begin to fail, inducing increased local stresses, which cause more bonds to fail.

Buckling: complicated, because there are many modes

1st, 2nd, and 3rd orderbending modes. Lowestorder is most likelyto occur.

Page 37: Materials – PowerPoint Presentation 2015

Euler buckling

http://upload.wikimedia.org/wikipedia/commons/thumb/6/69/Buckledmodel.JPG/1024px-Buckledmodel.JPG

How the specimen ends are supported must be considered.

Page 38: Materials – PowerPoint Presentation 2015

Material testing - Euler buckling load, PcPc load (MLT-2)

I moment of inertia (L4)

E Young’s modulus (ML-1T2)

L length (L)

We have 4 variables, 3 primitive dimensions = 1 dimensionless group

materials

Page 39: Materials – PowerPoint Presentation 2015

Euler buckling load

2

2

)(KL

EIF

The force at which a slender column under compression will failby bending

E = Young’s modulusI = area moment of inertiaL = unsupported length

K = 1.0 (pinned at both ends)= 0.699 (fixed at one end, pinned at the other= 0.5 (fixed at both ends)= 2.0 (free at one end, pinned at the other)

Page 40: Materials – PowerPoint Presentation 2015

I = area moment of inertia (dim L4)—associated with the bending of beams. Sometimes called second moment of area.

Not to be confused with

I = mass moment of inertia (dim ML2) — associated with the energy of rotation)

Area moment of inertia

Page 41: Materials – PowerPoint Presentation 2015

Some area moments of inertia

12

4aI

64

4dI

64

)( 44 dDI

12

3bdI

12

2 33 htsbI

Page 42: Materials – PowerPoint Presentation 2015

deflection y

load P

length L

Material Strength - Bending

compression: proportional

to distance from neutral axis

tension: proportional to

distance from neutral axis

neutral axis

shear

load

support

Page 43: Materials – PowerPoint Presentation 2015

y y dAs ( )

s smax maxy dA I2

But, s is proportional to strain e, and strain varies linearly with distance to the neutral line. Therefore, s = y smax , where smax is the stress at the maximum distance from the neutral line. So, Restoring moment =

where I is the area moment of inertia of the cross section of the beam about the neutral axis.

Moment of inertia depends on cross-section geometry and has units L4.

dAy

distance to

neutral line

s(y)

Restoring moment = (moment arm about neutral line) x (force) =

Page 44: Materials – PowerPoint Presentation 2015

Material testing - bendingmaterials

3-point Bend Test

Page 45: Materials – PowerPoint Presentation 2015

• Bending Test – Setup

Page 46: Materials – PowerPoint Presentation 2015

Materials good in compression

stone, concrete

Materials good in tension

carbon fiber, cotton, fiberglass

Materials good in both compression and tension

steel, wood

Page 47: Materials – PowerPoint Presentation 2015

Other strengths

Shear strength--rotating axles fail because their shear strengths were exceeded

Ultimate tensile strength--maximum possible load without failure

Yield strength--load required to cross line from elastic to plastic deformation

Page 48: Materials – PowerPoint Presentation 2015

NEXT: How do we put materials together to form structures. . .?

Page 49: Materials – PowerPoint Presentation 2015

Beams and loads--tension:

Beam under tension

Maximum load is tensile strength times cross-sectional area.Lmax = T * Acs

For regular spaghetti (diameter = 2mm), maximum load is ~ 10 pounds.

Load capacity does not depend on length.

Failure occurs when ultimate tensile strength is exceeded.

Page 50: Materials – PowerPoint Presentation 2015

Beams and loads--compression:

Beam in compression

Failure occurs two ways:

1) When L/d < 10, failure is by crushing

2) When L/d > 10, failure is by buckling

We are almost always concerned with failure by buckling.

L

d

Page 51: Materials – PowerPoint Presentation 2015

Buckling strength F = k * d4/L2

To determine constant of proportionality k:

1) Measure length and diameter of a piece of spaghetti2) Hold spaghetti vertically on postal scale3) Press down on spaghetti until it begins to bend4) Read load F on postal scale5) Calculate k

Beams and loads--compressive buckling:

Page 52: Materials – PowerPoint Presentation 2015

Some consequences of buckling properties:

If a beam of length L and diameter d can support acompressive load of F,

L

dF

then a beam of length L/2 and diameter d cansupport a compressive load of 4F.

L/2

d4F

_______?_______

__?___

Page 53: Materials – PowerPoint Presentation 2015

L

2d

16F

then a beam of length L and diameter 2d cansupport a compressive load of 16F.

L

dF

ALSO…

If a beam of length L and diameter d can support acompressive load of F,

__?__

_?__

Page 54: Materials – PowerPoint Presentation 2015

Beams and loads--bending:

Very little strength. Never design a structure thatrelies on bending strength to support a load.